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From a kinetics experiment, the Vmax was determined to be 525µM∙min-1. For the kinetic assay, 0.1mL...

From a kinetics experiment, the Vmax was determined to be 525µM∙min-1. For the kinetic assay, 0.1mL of a 0.25mg/mL solution of enzyme was used, and the enzyme has a molecular weight of 125,000 g/mole. Assume a reaction volume of 800µL. Calculate the kcat (in sec-1) for the enzyme. show all work!!!

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In this question we have to find Kcat. For this we can use formula of Vmax from the enzyme catalysis. But before that we have to find the concentration of enzyme. To find it i have used simple calculation by interpreting given data. First the weight of enzyme is found and then the concentration i. e molarity of the solution with respect to the enzyme is found.

Then we need to find the Kcat in per sec, so i have converted Vmax also in uM/s using fact that 1 min=60sec

Then both the values are put in the formula and the Kcat is found

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